On Sun, Dec 20, 2009 at 5:33 AM, Gabor Grothendieck
<ggrothendi...@gmail.com> wrote:
> Use a zero lookaround expression. It will not consume its match. See ?regexp
That should be lookahead, not lookaround.
>
>> gregexpr("a(?=a)", "aaa", perl = TRUE)
> [[1]]
> [1] 1 2
> attr(,"match.length")
> [1] 1 1
>
>
> On Sun, Dec 20, 2009 at 1:43 AM, Jonathan <jonsle...@gmail.com> wrote:
>> Last one for you guys:
>>
>> The command:
>>
>> length(gregexpr('cus','hocus pocus')[[1]])
>> [1] 2
>>
>> returns the number of times the substring 'cus' appears in 'hocus pocus'
>> (which is two)
>>
>> It's returning the number of **disjoint** matches. So:
>>
>> length(gregexpr('aa','aaa')[[1]])
>> [1] 1
>>
>> returns 1.
>>
>> **What I want to do:**
>> I'm looking for a way to count all occurrences of the substring, including
>> overlapping sets (so 'aa' would be found in 'aaa' two times, because the
>> middle 'a' gets counted twice).
>>
>> Any ideas would be much appreciated!!
>>
>> Signing off and thanks for all the great assistance,
>> Jonathan
>
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